\newproblem{lay:2_9_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.9.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Consider $A=\begin{pmatrix}1 & 3 & 2 & -6 \\ 3 & 9 & 1 & 5 \\ 2 & 6 & -1 & 9 \\ 5 & 15 & 0 & 14\end{pmatrix}$ and its echelon form
	$\begin{pmatrix}1 & 3 & 3 & 2 \\ 0 & 0 & 5 & -7 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0\end{pmatrix}$. Find bases for $\mathrm{Nul}\{A\}$ and
	$\mathrm{Col}\{A\}$.
}{
  % Solution
	The basis of $\mathrm{Nul}\{A\}$ is found by the equation system $A\mathbf{x}=\mathbf{0}$ whose augmented matrix is row-equivalent to
	\begin{center}
		$\left(\begin{array}{rrrr|r}1 & 3 & 3 & 2 & 0\\ 0 & 0 & 5 & -7 & 0\\ 0 & 0 & 0 & 5 & 0\\ 0 & 0 & 0 & 0&0\end{array}\right)$
	\end{center}
	We may calculate its reduced echelon form
	\begin{center}
		$\left(\begin{array}{rrrr|r}1 & 3 & 3 & 2 & 0\\ 0 & 0 & 5 & -7 & 0\\ 0 & 0 & 0 & 5 & 0\\ 0 & 0 & 0 & 0&0\end{array}\right) \sim
		 \left(\begin{array}{rrrr|r}1 & 3 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0&0\end{array}\right)=B$
	\end{center}
	This implies the following equations:
	\begin{center}
		$x_1=-3x_2$ \\
		$x_3=0$ \\
		$x_4=0$
	\end{center}
	So the basis of $\mathrm{Nul}\{A\}$ is given by the non-pivot columns of $B$, i.e.,
	\begin{center}
	   $\mathrm{Basis}\{\mathrm{Nul}\{A\}\}=\{(-3,1,0,0)\}$
	\end{center}
	The basis of 
	$\mathrm{Col}\{A\}$ is given by the pivot columns of $B$. The basis of the column space of $B$ is given by its first, third and fourth columns
	($\{\mathbf{b}_1,\mathbf{b}_3,\mathbf{b}_4\}$). Similarly, the basis of the column space of $A$ is given by its first, third and fourth columns
	($\{\mathbf{a}_1,\mathbf{a}_3,\mathbf{a}_4\}$), i.e.,
	\begin{center}
	   $\mathrm{Basis}\{\mathrm{Col}\{A\}\}=\{(1,3,2,5),(2,1,-1,0),(-6,5,9,14)\}$
	\end{center}
}
\useproblem{lay:2_9_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
